To what interval I must we restrict the parameter t if the graph of the parametric equations
x=t+1
y = 2t^2 +1
for t in the set I
is identical to the graph of the parametric equations
x = 1+ sin s
y = 2- cos(2s)
for s in [0,pi)?
it appears we need
1+t = 1+sin s,so
t = sin s
t must be in [0,1] if s is in [0,pi)
also,
2t^2+1 = 2-cos(2s)
2t^2+1 = 2 - (1-2sin^2 s)
2t^2+1 = 2sin^2 s + 1
same domain.
maybe you misunderstood
Thx!
@steveiswrong by same domain, he was talking about the domain of t, [0,1], and you misunderstood him. Also, please do not insult people by adding a wrong/right to their name. It's rude. The kind sir took his time to write that solution. You should be thanking him.
To determine the interval for the parameter t, we need to find the corresponding values of t for the given interval [0, pi) in the parametric equations x = t + 1 and y = 2t^2 + 1.
First, let's find t in terms of s by equating the two x-equations:
t + 1 = 1 + sin(s)
Next, let's solve this equation for t:
t = sin(s)
Now that we have t in terms of s, let's substitute it in the y-equation:
y = 2t^2 + 1
= 2(sin(s))^2 + 1
= 2sin^2(s) + 1
Now, we need to determine the interval for t that corresponds to the interval [0, pi) for s.
In the interval [0, pi), sin(s) ranges from 0 to 1. Therefore, substituting the values of sin(s) in the equation for t, we find that t will range from 0 to 1.
Hence, the interval for the parameter t is [0, 1].